Undergrad Determining the flux of an arbitrary vector function

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To determine if an arbitrary vector function f(v) has constant flux over a closed surface, one can analyze properties such as divergence and potential. The flux of f(v) across the surface is defined as a numerical value, which can be constant under certain conditions. Understanding the relationship between the function's characteristics and its flux is crucial for deriving the desired physical formula. The discussion emphasizes the need for clarity on what specific flux is being referenced. The exploration of these concepts is essential for successful application in vector calculus.
DarkBabylon
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Hello there. I've been working on trying to re-derive a certain physical formula using vector calculus, and came to a conclusion that in order to derive it, I'll need a way to determine the nature of a certain expression.
Specifically:
∯f(v)·da - v={x1,x2,x3,...,xn} and f(v) returns a vector in the same space of v.
Is there a way to determine if a certain arbitrary function f(v) has a constant Flux, as one would call it, on a closed surface using some other information about f(v) such as its potential (∮f(v)· dv), divergence, etc.?
 
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DarkBabylon said:
if a certain arbitrary function f(v) has a constant Flux,

What flux do you want to be constant? The flux of a given function over a given surface has some numerical value - i.e. it is a constant.
 
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Stephen Tashi said:
i.e. it is a constant.
Oh, right. o0) Thanks.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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